Question

A proton travels through uniform magnetic and electric fields. the magnetic field is in the negative x direction and has a magnitude of 2.04 mt. at one instant the velocity of the proton is in the positive y direction and has a magnitude of 1930 m/s. at that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 4.70 v/m, (b) in the negative z direction and has a magnitude of 4.70 v/m, and (c) in the positive x direction and has a magnitude of 4.70 v/m?

Answer 1

For a particle moving in the +y and in a -x B field the force will be in the + z direction Magnitude = q*v*B = 1.60 x 10^-19*1930*2.04 x 10^-3 = 0.63x10^-18N a) the electric field creates a force = E*q in the +z direction Fe = 4.27*1.60x10^-19 = 6.83 x 10^-19N So F = 0.63 x 10^-18 + 6.83 x 10^-19 = 1.31 x 10^-18N b) Now F = 0.63 x 10^-18 – 6.83 x 10^-19 = 0.5 x 10^-19N c) Now add them as vectors so F = sqrt((0.63 x 10^-18)^2 + (6.83 x 10^-19)^2) = 0.86 x 10^-18N